we can see that$$\frac d{dx}\ln(ax)=\frac1x$$so why is$$\int\frac{dx}{x}\neq\ln(ax)+C$$why can we omit the possible constant, a?

Question

callisto · Answer

i think you've omitted the constant a in derivative

campbell_st · Answer

they cancel each other... 

eg...  y = ln(3x)   dy/dx = 3/3x = 1/x

turingtest · Answer

yeah^ thank you

turingtest · Answer

every friggin' time they cancel, so why do we exclude it from the integral?

experimentx · Answer

i think you are adding two constants on second integration 'a' and 'C'
I think you can remove 'a' and write 'ln a' inplace of C, you will get the same result

turingtest · Answer

but$$\ln|ax|+C\neq\ln|x|+C$$if $a\neq0$

turingtest · Answer

I mean if $a\neq1$

mysesshou · Answer

my book has $$\frac{d}{dx}(ln (u)) = \frac{u'}{u}$$   ... so I guess if you use that one, then you would still have the a constant, when u=ax

Also, I have $$\frac{du}{u} = ln \left| u \right| +C$$  which would also show the a/ax where it also cancels.
where u=ax
But, yeah, that is interesting.

experimentx · Answer

∫dxx=ln(x)+C , since 'C' is a constant, you can equivalently write 'ln(a)' for it, and you get ln(ax) which is again the general form

turingtest · Answer

now that makes sense!
thank you @experimentX

experimentx · Answer

you are welcome

mysesshou · Answer

That does make sense for C=ln(a)  ... Never really thought about the why's lately   :)